0
이 문제와 관련하여 이전 게시물이 있지만 문제가 해결되지 않았습니다. 배포 중 오류 - EntityManagerFactory를 빌드 할 수 없습니다.
내가 응용 전쟁을 배포
, 나는 다음과 같은 오류Invocation of init method failed; nested exception is javax.persistence.
PersistenceException: [PersistenceUnit: default] Unable to build EntityManagerFactory
at oracle.oc4j.admin.internal.DeployerBase.execute(DeployerBase.java:133)
at
oracle.oc4j.admin.jmx.server.mbeans.deploy.OC4JDeployerRunnable.doRun
(OC4JDeployerRunnable.java:52)
at oracle.oc4j.admin.jmx.server.mbeans.deploy.DeployerRunnable.run
(DeployerRunnable.java:81)
at com.evermind.util.ReleasableResourcePooledExecutor$MyWorker.run
(ReleasableResourcePooledExecutor.java:298)
at java.lang.Thread.run(Unknown Source)
이 문제를 해결하는 방법을 잘하지를 얻고있다.
ApplicationContext.xml
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/tx
http://www.springframework.org/schema/tx/spring-tx-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd"
>
<!-- Data Source Declaration -->
<bean id="DataSource" class="org.springframework.jndi.JndiObjectFactoryBean">
<property name="jndiName" value="jdbc/scottDS"/>
</bean>
<context:component-scan base-package="net.test" />
<bean class="org.springframework.dao.annotation.PersistenceExceptionTranslationPostProcessor"/>
<bean id="entityManagerFactory"
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="dataSource" ref="DataSource" />
<property name="packagesToScan" value="net.test" />
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
<property name="showSql" value="true" />
<property name="generateDdl" value="false" />
<property name="databasePlatform" value="${jdbc.dialectClass}" />
</bean>
</property>
</bean>
<bean id="defaultLobHandler" class="org.springframework.jdbc.support.lob.DefaultLobHandler" />
<!-- Transaction Config -->
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
<tx:annotation-driven transaction-manager="transactionManager"/>
<context:annotation-config/>
</beans>
엔티티 클래스
EMP를
@Entity
@Table(name = "EMP"
)
@XmlRootElement
public class Emp implements java.io.Serializable {
private short empno;
private Dept dept;
private String ename;
private String job;
private Short mgr;
private Date hiredate;
private BigDecimal sal;
private BigDecimal comm;
public Emp() {
}
public Emp(short empno) {
this.empno = empno;
}
public Emp(short empno, Dept dept, String ename, String job, Short mgr, Date hiredate,
BigDecimal sal, BigDecimal comm) {
this.empno = empno;
this.dept = dept;
this.ename = ename;
this.job = job;
this.mgr = mgr;
this.hiredate = hiredate;
this.sal = sal;
this.comm = comm;
}
@Id
@Column(name = "EMPNO", unique = true, nullable = false, precision = 4, scale = 0)
public short getEmpno() {
return this.empno;
}
public void setEmpno(short empno) {
this.empno = empno;
}
@ManyToOne
@JoinColumn(name="DEPTNO")
public Dept getDept() {
return this.dept;
}
public void setDept(Dept dept) {
this.dept = dept;
}
@Column(name = "ENAME", length = 10)
public String getEname() {
return this.ename;
}
public void setEname(String ename) {
this.ename = ename;
}
@Column(name = "JOB", length = 9)
public String getJob() {
return this.job;
}
public void setJob(String job) {
this.job = job;
}
@Column(name="MGR", precision=4, scale=0)
public Short getMgr() {
return this.mgr;
}
public void setMgr(Short mgr) {
this.mgr = mgr;
}
@Temporal(TemporalType.DATE)
@Column(name="HIREDATE", length=7)
public Date getHiredate() {
return this.hiredate;
}
public void setHiredate(Date hiredate) {
this.hiredate = hiredate;
}
@Column(name="SAL", precision=7)
public BigDecimal getSal() {
return this.sal;
}
public void setSal(BigDecimal sal) {
this.sal = sal;
}
@Column(name="COMM", precision=7)
public BigDecimal getComm() {
return this.comm;
}
public void setComm(BigDecimal comm) {
this.comm = comm;
}
}
부서
@Entity
@Table(name="DEPT"
)@XmlRootElement
public class Dept implements java.io.Serializable {
private byte deptno;
private String dname;
private String loc;
private Set emps = new HashSet(0);
public Dept() {
}
public Dept(byte deptno) {
this.deptno = deptno;
}
public Dept(byte deptno, String dname, String loc,Set emps) {
this.deptno = deptno;
this.dname = dname;
this.loc = loc;
this.emps = emps;
}
@Id
@Column(name="DEPTNO", unique=true, nullable=false, precision=2, scale=0)
public byte getDeptno() {
return this.deptno;
}
public void setDeptno(byte deptno) {
this.deptno = deptno;
}
@Column(name="DNAME", length=14)
public String getDname() {
return this.dname;
}
public void setDname(String dname) {
this.dname = dname;
}
@Column(name="LOC", length=13)
public String getLoc() {
return this.loc;
}
public void setLoc(String loc) {
this.loc = loc;
}
@OneToMany(mappedBy="dept")
public Set<Emp> getEmps() {
return this.emps;
}
public void setEmps(Set emps) {
this.emps = emps;
}
}
업데이트
의 persistence.xml은
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/tx
http://www.springframework.org/schema/tx/spring-tx-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd"
>
<context:component-scan base-package="net.test" />
<bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="persistenceUnitName" value="default" />
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
<property name="generateDdl" value="false" />
<property name="showSql" value="true" />
<property name="databasePlatform" value="org.hibernate.dialect.OracleDialect" />
</bean>
</property>
</bean>
<!-- Transaction Config -->
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
<tx:annotation-driven transaction-manager="transactionManager"/>
<context:annotation-config/>
</beans>
applicationContext.xml
수정<persistence xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence persistence_1_0.xsd" version="1.0">
<persistence-unit name="default" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<class>net.test.entity.Emp</class>
<class>net.test.entity.Dept</class>
<properties>
<property name="toplink.logging.level" value="FINE"/>
<property name="toplink.jdbc.driver" value="oracle.jdbc.OracleDriver"/>
<property name="toplink.jdbc.url" value="jdbc:oracle:thin:@serverdb:1521:devp"/>
<property name="toplink.jdbc.password" value="scott"/>
<property name="toplink.jdbc.user" value="tiger"/>
</properties>
</persistence-unit>
</persistence>
업데이트 2
당신은 상단 링크를 혼합 최대 절전 모드 것 같다
<persistence xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence persistence_1_0.xsd" version="1.0">
<persistence-unit name="default" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<class>net.test.entity.Emp</class>
<class>net.test.entity.Dept</class>
<properties>
<property name="hibernate.connection.driver_class" value="oracle.jdbc.OracleDriver" />
<property name="hibernate.connection.url" value="jdbc:oracle:thin:@serverdb:1521:DEVP" />
<property name="hibernate.connection.username" value="scott" />
<property name="hibernate.connection.password" value="tiger" />
<property name="hibernate.show_sql" value="true" />
</properties>
</persistence-unit>
</persistence>
영구 단위를 'default'라는 이름으로 정의 했습니까? – Gaurav
@Gaurav default라는 영구적 유닛을 찾을 수 없습니다. 응용 프로그램을 배포 할 때 응용 프로그램 서버는 부모 응용 프로그램을 요청하고 기본값은 드롭 다운 선택입니다. 이로 인해 문제가 발생합니까? – user75ponic
persistence.xml을 게시하는 것이 다소 의미가 있습니까? 당신이 거기에 퍼시스턴스 유닛을 정의했기 때문에 사람들은 그것을 볼 수 없기 때문에 (그리고 JPA 역시 그것을 볼 수 없다.) –